Power Developed by Synchronous Motor

In this chapter, we will derive the expression for mechanical power developed (Pm) by a three-phase synchronous motor. Here, we will neglect the armature resistance Ra of the synchronous motor. Then, the armature copper loss will be zero and hence the mechanical power developed by the motor is equal to the input power (Pin) to the motor, i.e.,

$$\mathrm{\mathit{P_{m}}\:=\:\mathit{P_{in}}}$$

Now, consider an under-excited (i.e.,Eb<V) three-phase synchronous motor having zero armature resistance (i.e.,Ra = 0), and is driving a mechanical load.

Mechanicalpower

The phasor diagram of one phase of this synchronous motor is shown in the figure. Because the motor is under-excited, thus it will operate at a lagging power factor, say (cos $\phi$). From the phasor diagram, it is clear that $\mathit{E_{r}}\:=\:I_{a}X_{s}$ and the armature current per phase $I_{a}$ lags the resultant EMF $\mathit{E_{r}}$ by an angle of 90°.

Therefore, the input power per phase to the motor is given by,

$$\mathrm{\mathit{P_{in}}\:=\mathit{VI_{a}}\:cos\:\phi \:\cdot \cdot \cdot (1)}$$

Since $\mathit{P_{m}}$ is equal to $\mathit{P_{in}}$, therefore,

$$\mathrm{\mathit{P_{m}}\:=\mathit{VI_{a}}\:cos\:\phi \:\cdot \cdot \cdot (2)}$$

From the phasor diagram, we have,

$$\mathrm{\mathit{AB}\:=\mathit{I_{a}X_{s}}\:cos\:\phi \:=\:\mathit{E}_{\mathit{b}}\:sin\:\delta}$$

$$\mathrm{\therefore \mathit{I_{a}\:cos\phi \:=\:\frac{E_{b}\:sin\delta }{\mathit{X_{s}}}}\:\cdot \cdot \cdot (3)}$$

Using equations (2) & (3), we obtain,

$$\mathit{P_{m}\:=\:\frac{\mathit{VE_{b}}sin\delta }{X_{s}}}\cdot \cdot \cdot (4)$$

This is the expression for mechanical power developed (Pm) per phase by the synchronous motor.

For 3-phases of the motor, the developed mechanical power is given by,

$$\mathit{P_{m}\:=\:\frac{3\mathit{VE_{b}}sin\delta }{X_{s}}}\cdot \cdot \cdot (5)$$

Also, from the equations (4) & (5), it is clear that the mechanical power developed will be maximum when power angle ($\delta$= 90°) electrical. Therefore,

For per phase,

$$\mathit{P_{max}\:=\:\frac{\mathit{VE_{b}}}{X_{s}}}\cdot \cdot \cdot (6)$$

For 3-phase,

$$\mathit{P_{max}\:=\:\frac{\mathit{3VE_{b}}}{X_{s}}}\cdot \cdot \cdot (7)$$

Important Points

The following important points may be noted about the mechanical power developed by a three-phase synchronous motor −

  • The mechanical power developed by the synchronous motor increases with the increase in power angle ($\delta$) and vice-versa.

  • If power angle ($\delta$) is zero, then the synchronous motor cannot develop mechanical power.

  • When the field excitation of the synchronous motor is reduced to zero, i.e., ($E_{b}$ = 0), the mechanical power developed by the motor is also zero, i.e. the motor will come to a stop.

Numerical Example

A 3-phase, 4000 kW, 3.3 kV, 200 RPM, 50 Hz synchronous motor has per phase synchronous reactance of 1.5 $\Omega$. At full-load, the power angle is 22° electrical. If the generated back EMF per phase is 1.7 kV, calculate the mechanical power developed. What will be the maximum mechanical power developed?

Solution

Given data,

  • Voltage per phase, $\mathit{V}\:=\:\frac{3.3}{\sqrt{3}}\:=\:1.9\:kV$

  • Back EMF per phase,$\mathit{E_{b}}\:=\:1.7\:kV$

  • Synchronous reactance,$X_{s}\:=\:1.5\Omega $

  • Power angle,$\delta \:=\:22^{^{\circ}}$

Therefore, the mechanical power developed by the motor is,

$$\mathrm{\mathit{P_{m}}\:=\:\frac{3\:\mathit{VE_{b}\:sin\delta} }{\mathit{X_{s}}}\:=\:\frac{3\times 1.9\times 1.7\times \mathrm{sin}\:22^{\circ}}{1.5}}$$

$$\mathit{\therefore P_{m}}\:=\:2.42\times 10^{6}\:W\:=\:2.42\:MW$$

The mechanical power developed will be maximum when ($\delta$ =90°),

$$\mathit{P_{max}}\:=\:\frac{3\mathit{VE_{b}}}{X_{s}}\:=\:\frac{3\times 1.9\times 1.7}{1.5}$$

$$\mathit{\therefore P_{max}}\:=\:6.46\:\mathrm{MW}$$

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